Learning Zig

Explain
1

Hello guys, this is my first time learning Zig and programming. Why I am getting stuck at while loop in this code below if I typing let say “Jack”, sorry If my question doesn’t make sense and sorry my English is not that good.

const std = @import("std");
const File = std.Io.File;

pub fn main(init: std.process.Init) !void {
    const io = init.io;

    var stdout_buffer: [4096]u8 = undefined;
    var stdout_writer = File.stdout().writer(io, &stdout_buffer);
    const stdout = &stdout_writer.interface;

    var stdin_buffer: [4096]u8 = undefined;
    var stdin_reader = File.stdin().reader(io, &stdin_buffer);
    const stdin = &stdin_reader.interface;

    while (true) {
        try stdout.print("Enter your name: ", .{});
        try stdout.flush();

        const line = try stdin.takeDelimiterExclusive('\n');
        const name = std.mem.trim(u8, line, "\r\n");

        if (std.mem.eql(u8, name, "John")) {
            try stdout.print("My name is {s}\n", .{name});
            try stdout.flush();
            break;
        }
    }
}
2

You’re using takeDelimiterExclusive, so what you’re taking never includes the delimiter (it’s excluded, per the name of the method). So, your reading never progresses past the first delimiter: it is never “taken”.

You’ll want to use takeDelimiterInclusive (depending on Zig version these may be named differently, such as takeDelimiter). This also renders doing the trimming sensible.

3

Whoa, okay that’s work really well.

I have question if you don’t mind. The delimiter is “\n” right? so, let say I typing “Jack” if I’m using takeDelimiterInclusiveI should get the value “Jack\n” right? but why if I’m using takeDelimiterExclusive I got the value of empty string ““?

Thank you for the reply!